how to prove $G$ is an abelian group under $*$ (called the real numbers mod 1) Let $G = \{x \in \mathbb{R}~|~0\leq x < 1\}$ and for $x,y \in G$ let $x*y$ be the fractional part of $x+y$ i.e $x*y = x + y - [x + y]$ where $[a]$ is the greatest integer less than or equal to $a$. I need help proving $*$ is a well defined binary operation on $G$ and that $G$ is an abelian group under $*$.
To prove that $*$ is well defined, I rely on the assumption that +,- is well defined in the set of real numbers. evaluating the brackets [] is also well-defined in the set of real numbers. (is this an okay assumption?)
To prove $*$ is associative, (I imagine because +,- is associative, so will *?) I show that $(x*y)*z = x*(y*z)$ $$(x*y)*z =(x+y-[x+y])*z = x+y-[x+y]+z - [x+y-[x+y]+z]$$ $$x*(y*z) = x*(y+z - [y+z]) = x + y + z - [y+z] - [x+y+z - [y+z]]$$ rearranging $$(x*y)*z = x+y+z-[x+y]-[x+y+z-[x+y]]$$ $$x*(y*z) = x+y+z-[y+z]-[x+y+z-[y+z]]$$ I am not so sure that these two are equal. I don't see how distributing the $-$ to remove the terms $[x+y]$ and $[y+z]$ is fair (ex: $x = y = \frac{1}{2}$ so $x*y = .5+.5-[.5]-[.5]$ yields a different answer than $x*y = .5+.5-[.5+.5]$
The identity element would be $0$, the inverse would have to be $-x$ which isn't in $G$. What am I doing wrong?
edit: commutativity would be proven by showing $x*y = y*x$ which is easy to show $x+y - [x+y] = y+x - [y+x]$ right?
 A: 
To prove $∗$ is associative

Let $a,b,c\in G$.
\begin{align*}
(a*b)*c &= (a+b-[a+b])*c \\
   &= a+b+c-[a+b]-\left[a+b+c-[a+b]\right]
\end{align*}
Similarly,
\begin{align*}
a*(b*c) &= a+b+c-[b+c]-\left[a+b+c-[b+c]\right]
\end{align*}
Given $0\le a,b,c < 1, [a+b]$ and $[b+c]$ must be either zero or one individually, for a total of four possible cases.  In the cases when both are zeros or ones, clearly $(a*b)*c = a*(b*c)$.
Consider the case when $[a+b]=1$ and $[b+c]=0$.  Necessarily $1 \le a+b < 2, b+c< 1$ and $1\le a+b+c< 2\implies [a+b+c] = 1$ and $[a+b+c-1] = 0$. Therefore, $(a*b)* c=a+b+c-1-[a+b+c-1]=a+b+c-1$, whereas $a*(b* c)=a+b+c-[a+b+c]=a+b+c-1$; i.e. $(a* b)* c = a*(b* c)$.
Similar argument applies when $[a+b]=0$ and $[b+c]=1$.  This covers all four cases.
Hence $G$ is associative over $*$.
A: You can think of your group $(G, *)$ as $(\mathbb{R}, +)/\mathbb{Z}$: That is, the reals under addition modded out by the integers. 
This clearly works because we are essentially saying that is $x, y \in \mathbb{R}$, $x - y \in \mathbb{Z}$, then $x = \mathbb{Z} + y$ in $\mathbb{R} /\mathbb{Z}$.
This has the same effect as “truncation”, but is now a nice subgroup
Hence, 
$$
(G, *) \simeq (\mathbb{R}/\mathbb{Z}, +)
$$
To imagine this, realize that $\mathbb{R}$ just contains shifted copies of $[0, 1)$ all laid down one next to the other. So, we simply “collapse” all these copies together for this construction to work out.
Now, for this to be a group, we need the kernel of the map $$\phi: (\mathbb{R}, +) \to (\mathbb{R}, +) /\mathbb{Z}$$
to be a normal subgroup. Since the kernel of a quotient map is the quotienting subgroup itself, we simply need to
show that $\mathbb{Z}$ is a normal subgroup of $\mathbb{R}$. This is obvious in abelian groups.
